Problem: $h(t) = -2t$ $f(t) = -5t^{3}+6t^{2}+6t+2(h(t))$ $ f(h(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = (-2)(-1)$ $h(-1) = 2$ Now we know that $h(-1) = 2$ . Let's solve for $f(h(-1))$ , which is $f(2)$ $f(2) = -5(2^{3})+6(2^{2})+(6)(2)+2(h(2))$ To solve for the value of $f$ , we need to solve for the value of $h(2)$ $h(2) = (-2)(2)$ $h(2) = -4$ That means $f(2) = -5(2^{3})+6(2^{2})+(6)(2)+(2)(-4)$ $f(2) = -12$